Summable series and the Riemann rearrangement theorem Jordan Bell [email protected] Department of Mathematics, University of Toronto May 4, 2015 1 Introduction Let N be the set of positive integers. A function from N to a set is called a sequence. If X is a topological space and x ∈ X, a sequence a : N → X is said to converge to x if for every open neighborhood U of x there is some NU such that n ≥ NU implies that an ∈ U . If there is no x ∈ X for which a converges to x, we say that a diverges. Pn Let a : N → R. We define s(a) : N → R by sn (a) = k=1 ak . We call sn (a) the nth partial sum of the sequence a, and we call the sequence s(a) a series. If there is some σ ∈ R such that s(a) converges to σ, we write ∞ X ak = σ. k=1 2 Goldbach Euler [22, §110]: “If, as is commonly the case, we take the sum of a series to be the aggregate of all of its terms, actually taken together, then there is no doubt that only infinite series that converge continually closer to some value, the more terms we actually add, can have sums”. Euler Goldbach correspondence nos. 55, 161, 162. 3 Dirichlet In 1837 Dirichlet proved that one can rearrange terms in an absolutely convergent series and not change the sum, and gave examples to show that this was not the case for conditionally convergent series. If a is a sequence and the series s(|a|) converges, we say that the series s(a) is absolutely convergent. Because R is a complete metric space, a series being absolutely convergent implies that it is convergent. 1 Dirichlet [14] and [15, p. 176, §101] Elstrodt [20] and [19] In the following theorem we prove that if a series converges absolutely, then every rearrangement of it converges to the same value. Our proof follows Landau [45, p. 157, Theorem 216]. Theorem 1. If a is a sequence for which s(a) converges absolutely and ∞ X an = σ, n=1 then for any bijection λ : N → N, the series s(a ◦ λ) converges to σ. Proof. Let > 0, and let M be large enough so that ∞ X |an | < . n=M Let r be large enough so that {n : 1 ≤ n < M } ⊆ {λn : 1 ≤ n ≤ r}. Fix m ≥ r, and let h : N → N be the sequence whose terms are the elements of N \ {λn : 1 ≤ n ≤ m} arranged in ascending order. If t + m ≥ max1≤n≤m λn then {λn : 1 ≤ n ≤ m} ∪ {hn : 1 ≤ n ≤ t} = {n : 1 ≤ n ≤ t + m}, and hence m X aλn + n=1 t X ahn = n=1 Taking t → ∞, we get m X aλn + n=1 t+m X an . n=1 ∞ X ahn = σ; n=1 the series s(a ◦ h) converges because for sufficiently large n, hn = n. Hence, for every m ≥ r, m ∞ ∞ ∞ X X X X aλn − σ = ahn ≤ |ahn | ≤ |an | < δ, n=1 n=1 n=1 which shows that s(a ◦ λ) converges to σ. 2 n=M 4 Riemann rearrangement theorem If a : N → R and λ : N → N is a bijection, we call the sequence a ◦ λ : N → R a rearrangement of the sequence a. Because N is a well-ordered set, if there are at least n elements in the set {k ∈ N : ak ≥ 0} then it makes sense to talk about the nth nonnegative term in the sequence a. If a were not a function from N to R but merely a function from a countable set to R, it would not make sense to talk about the nth nonnegative term in a or the nth negative term in a. Riemann [60, pp. 96-97] Our proof follows Landau [45, p. 158, Theorem 217]. Theorem 2 (Riemann rearrangement theorem). If a : N → R and s(a) converges but s(|a|) diverges, then for any nonnegative real number σ there is some rearrangement b of a such that s(b) → σ. Proof. Define p, q : N → R by pn = |an | + an , 2 qn = |an | − an . 2 pn and qn are nonnegative, and satisfy pn − qn = an , pn + qn = |an |. If one of s(p) or s(q) converges and the other diverges, we obtain a contradiction from sn (a) = n X k=1 ak = n X (pk − qk ) = k=1 n X pk − k=1 n X qk = sn (p) − sn (q) k=1 and the fact that s(a) converges. If both s(p) and s(q) converge, then we obtain a contradiction from sn (|a|) = n X k=1 |ak | = n X (pk + qk ) = n X pk + k=1 k=1 n X qk = sn (p) + sn (q) k=1 and the fact that s(|a|) diverges. Therefore, both s(p) and s(q) diverge. Because s(a) converges and s(|a|) diverges, there are infinitely many n with an > 0 and there are infinitely many n with an < 0. Let Pn be the nth nonnegative term in the sequence a, and let Qn be the absolute value of the nth negative term in the sequence a. The fact that s(p) diverges implies that s(P ) diverges, and the fact that s(q) diverges implies that s(Q) diverges. Let σ ≥ 0. We define sequences µ, ν : N → N by induction as follows. Let µ1 be the least element of N such that sµ1 (P ) > σ, and with µ1 chosen, let ν1 be the least element of N such that sµ1 (P ) − sν1 (Q) < σ. 3 Let m2 be the least element of N such that sµ2 (P ) − sν1 (Q) > σ, and with µ2 chosen, let ν2 be the least element of N such that sµ2 (P ) − sν2 (Q) < σ. It is straightforward to check that µ2 > µ1 and ν2 > ν1 . Suppose that µ1 , . . . , µn and ν1 , . . . , νn have been chosen, that µn is the least element of N such that sµn (P ) − sνn−1 (Q) > σ, that νn it the least element of N such that sµn (P ) − sνn (Q) < σ and that µn > µn−1 and νn > νn−1 . Let µn+1 be the least element of N such that sµn+1 (P ) − sνn (Q) > σ, and with µn+1 chosen, let νn+1 be the least element of N such that sµn+1 (P ) − sνn+1 (Q) < σ. It is straightforward to check that µn+1 > µn and νn+1 > νn . Define b : N → R by taking bn to be the nth term in P1 , . . . , Pµ1 , −Q1 , . . . , −Qν1 , Pµ1 +1 , . . . , Pµ2 , −Qν1 +1 , . . . , −Qν2 , . . . , which, because the sequences µ and ν are strictly increasing, is a rearrangement of the sequence a. 5 Symmetry Don’t use order where it is accidental. 6 Nets A directed set is a set D and a binary relation satisfying • if m, n, p ∈ D, m n, and n p, then m p • if m ∈ D, then m m • if m, n ∈ D, then there is some p ∈ D such that m p and n p. 4 For example, let A be a set, let D be the set of all subsets of A, and say that F G when F ⊆ G. Check that (D, ) is a directed set: for F, G ∈ D, we have F ∪ G ∈ D, and F ∪ G is an upper bound for both F and G. A net is a function from a directed set (D, ) to a set X. Let (X, τ ) be a topological space, let S : (D, ) → (X, τ ) be a net, and let x ∈ X. We say that S converges to x if for every U ∈ τ with x ∈ U there is some NU ∈ D such that NU i implies that S(i) ∈ U . One proves that a topological space is Hausdorff if and only if every net in this space converges to at most one point [41, p. 67, Theorem 3]. A net S : (D, ) → R is said to be increasing if m n implies that S(m) ≤ S(n). Lemma 3. If S : (D, ) → R is an increasing net and the range R of S has an upper bound, then S converges to the supremum of R. Proof. Because R is a subset of R that has an upper bound, it has a supremum, call it σ. To say that σ is the supremum of R means that for all r ∈ R we have r ≤ σ (σ is an upper bound) and that for all > 0 there is some r ∈ R with σ − < r (nothing less than σ is an upper bound). Take > 0. There is some r ∈ R with σ − < r . As r ∈ R, there is some n ∈ D with S(n ) = r . If n n, then because S is increasing, S(n ) ≤ S(n), and hence σ − < r = S(n ) ≤ S(n). But S(n) ∈ R, so S(n) ≤ σ. Hence n n implies that |S(n) − σ| < , showing that S converges to σ. 7 Unordered sums Let A be a set, and let P0 (A) be the set of all finite subsets of A. Check that (P0 (A), ⊆) is a directed set: if F, G ∈ P0 (A) then F ∪ G ∈ P0 (A) and F ∪ G is an upper bound for both F and G. Let f : A → R be a function, and define Sf : P0 (A) → R by X Sf (F ) = f (a), F ∈ P0 (A). a∈F If the net Sf converges, we say that the function f is summable, and we call the element of R to which Sf converges the unordered sum of f , denoted by X f (a). a∈A If B is a subset of A, we say that f is summable over B if the restriction of f to B is summable. If fB is the restriction of f to B and f is summable over B (i.e. fB is summable), by X f (a) a∈B 5 we mean X fB (a). a∈B Lemma 4. Suppose that f, g : A → R are functions and α, β ∈ R. If f and g are summable, then αf + βg is summable and X X X (αf (a) + g(a)) = α f (a) + β g(a). a∈A a∈A a∈A P P Proof. Let σ1 = a∈A f (a) and σ2 = a∈A g(a), and set h = αf + βg. For > 0, there is some F ∈ P0 (A) such that F ⊆ F ∈ P0 (A) implies that |Sf (F ) − σ1 | < , and there is some G ∈ P0 (A) such that G ⊆ G ∈ P0 (A) implies that |Sg (G) − σ2 | < . Let H = F ∪ G ∈ P0 (A). If H ⊆ H ∈ P0 (A), then, as F ⊆ H and G ⊆ H, X (αf (a) + βg(a)) − ασ1 − βσ2 |Sh (H) − (ασ1 + βσ2 )| = a∈H = |αSf (H) + βSg (H) − ασ1 − βσ2 | ≤ |α||Sf (H) − σ1 | + |β||Sg (H) − σ2 | ≤ |α| + |β|; we write ≤ rather than < in the last inequality to cover the case where α = β = 0. It follows that Sh converges to ασ1 + βσ2 . The following lemma is simple to prove and ought to be true, but should not to be called obvious. For example, the Ces`aro sum of the sequence 1, −1, 1, −1, . . . is 12 , while the Ces` aro sum of the sequence 1, −1, 0, 1, −1, 0, . . . is 31 . Lemma 5. If f : A → R is summable, then for any set C that contains A, the function g : C → R defined by ( f (c) c ∈ A g(c) = 0 otherwise is summable, and X f (a) = a∈A X g(c). c∈C P Proof. Let σ = a∈A f (a). For > 0, there is some F ∈ P0 (A) such that F ⊆ F ∈ P0 (A) implies that |Sf (F ) − σ| < . If F ⊆ H ∈ P0 (C), then, as 6 F ⊆ H ∩ A ∈ P0 (A), |Sg (H) − σ| = = = X g(c) − σ c∈H X X g(c) + g(c) − σ c∈H∩A c∈H\A X X f (a) + 0 − σ a∈H∩A c∈H\A = |Sf (H ∩ A) − σ| < . This shows that Sg converges to σ. The previous two lemmas are useful, and also convince us that unordered summation works similarly to finite sums. We now establish conditions under which a function is summable. Lemma 6. If f : A → R is nonnegative and there is some M ∈ R such that F ∈ P0 (A) for all Sf (F ) ≤ M P , then f is summable. If f : A → R is nonnegative and summable, then Sf (F ) ≤ a∈A f (a) for all F ∈ P0 (A). Proof. Suppose there is some M ∈ R such that if F ∈ P0 (A) then Sf (F ) ≤ M . That is, M is an upper bound for the range of Sf . Because f is nonnegative, the net Sf is increasing. We apply Lemma 3, which tells us that Sf converges to the supremum of its range. That Sf converges P means that f is summable. Suppose that f is summable, and let σ = a∈A f (a). Suppose by contradiction that there is some F0 ∈ P0 (A) such that Sf (F0 ) > σ, and let = Sf (F0 )−σ. Then there is some F ∈ P0 (A) such that F ⊆ F ∈ P0 (A) implies that |Sf (F ) − σ| < . As F ⊆ F0 ∪ F ∈ P0 (A), we have |Sf (F0 ∪ F ) − σ| < , and hence Sf (F0 ∪ F ) < σ + = Sf (F0 ). But F0 is contained in F0 ∪ F and f is nonnegative, so Sf (F0 ) ≤ Sf (F0 ∪ F ), which gives Sf (F0 ) < Sf (F0 ), a contradiction. Therefore, there is no F ∈ P0 (A) for which Sf (F0 ) > σ. Lemma 7. Suppose that f : A → R is a function and that A+ = {a ∈ A : f (a) ≥ 0} and A− = {a ∈ A : f (a) ≤ 0}. Then, f is summable if and only if f is summable over both A+ and A− . If f is summable, then X X X f (a) = f (a) + f (a). a∈A a∈A+ a∈A− 7 Proof. Suppose that f is summable. Because f is summable, there is some E ∈ P0 (A) such that E ⊆ F ∈ P0 (A) implies that |Sf (F ) − σ| < 1. Define E+ = {a ∈ E : f (a) ≥ 0} ∈ P0 (A+ ), E− = {a ∈ E : f (a) ≤ 0} ∈ P0 (A− ). If G ∈ P0 (A+ ) then E ⊆ G ∪ E ∈ P0 (A), and hence |Sf (G ∪ E) − σ| < 1. We have X X X X Sf+ (G) = f (a) ≤ f (a) = f (a) − f (a), a∈G a∈G∪E+ a∈G∪E a∈E− and hence Sf+ (G) ≤ Sf (G ∪ E) − Sf (E− ) < σ + 1 − Sf (E− ). That is, σ + 1 − Sf (E− ) is an upper bound for the range of Sf+ . The net Sf+ is increasing, hence applying Lemma 3 we get that Sf+ converges. That is, f+ is summable. If H ∈ P0 (A− ), then E ⊆ H ∪ E ∈ P0 (A), and hence |Sf (H ∪ E) − σ| < 1. We have X X X X Sf− (H) = f (a) ≥ f (a) = f (a) − f (a), a∈H a∈H∪E− a∈H∪E a∈E+ and then Sf− (H) ≥ Sf (H ∪ E) − Sf (E+ ) > σ − 1 − Sf (E+ ), showing that −σ + 1 + Sf (E+ ) is an upper bound for the net −Sf− . As −Sf− is increasing, by Lemma 3 it converges, and it follows that Sf− converges. That is, f− is summable. Suppose that f is summable over both A+ and A− . Let f+ be the restriction of f to A+ and let f+ be the restriction of f to A+ , and define g+ , g− : A → R by ( ( f (a) a ∈ A+ 0 a ∈ A+ g+ (a) = g− (a) = 0 a ∈ A− , f (a) a ∈ A− . By Lemma 5, f+ being summable implies that g+ is summable, with X X f+ (a) = g+ (a), a∈A+ a∈A and f− being summable implies that g− is summable, with X X f− (a) = g− (a). a∈A− a∈A But f = g+ + g− , so by Lemma 4 we get that f is summable, with X X X X X f (a) = g+ (a) + g− (a) = f+ (a) + f− (a). a∈A a∈A a∈A a∈A+ 8 a∈A− If f : A → R is a function, we define |f | : A → R by |f |(a) = |f (a)|. Theorem 8. If f : A → R is a function, then f is summable if and only if |f | is summable. Proof. Let A+ = {a ∈ A : f (a) ≥ 0} and A− = {a ∈ A : f (a) ≤ 0}, and let f+ and f− be the restrictions of f to A+ and A− respectively. Suppose that f is summable. Then by Lemma 7 we get that f+ is summable and f− is summable. Let F ∈ P0 (A) and write F+ = {a ∈ F : f (a) ≥ 0}, F− = {a ∈ F : f (a) ≤ 0}. We have X X X S|f | (F ) = |f (a)| = f (a) − f (a) = Sf+ (F+ ) − Sf− (F− ). a∈F a∈F+ a∈F− P But by Lemma 6, because the net Sf+ is increasing we have Sf+ (F+ ) ≤ a∈A+ f+ (a), P and because the net −Sf− is increasing we have −Sf− (F− ) ≤ − a∈A− f− (a). P P Therefore, a∈A+ f+ (a)− a∈A− f− (a) is an upper bound for the range of S|f | . Moreover, S|f | is increasing, so by Lemma 6 it follows that S|f | converges, i.e. that |f | is summable. Suppose that |f | is summable. By Lemma 6, for any F ∈ P0 (A+ ) we have X Sf+ (F ) = S|f | (F ) ≤ |f |(a), a∈A P i.e., a∈A |f |(a) is an upper bound for the range of Sf+ . As Sf+ is increasing, by Lemma 6 it follows that Sf+ converges, i.e., that f+ is summable. Because −Sf− is increasing, we likewise get that −Sf− converges and hence that Sf− converges, i.e. that f− is summable. Now applying Lemma 7, we get that f is summable. Theorem 9. If f : A → R is summable, then {a ∈ A : f (a) 6= 0} is countable. Proof. Suppose by contradiction that {a ∈ A : f (a) 6= 0} is uncountable. We have [ 1 {a ∈ A : f (a) 6= 0} = {a ∈ A : |f (a)| > 0} = a ∈ A : |f (a)| ≥ . n n∈N Since this is a countable union, there is some n ∈ N such that a ∈ A : |f (a)| ≥ n1 is uncountable; in particular, this set is infinite. Because f is summable, by Theorem 8 we have that |f | is summable, with unordered sum σ. Hence, there is some F1 ∈ P0 (A) such that F1 ⊆ F ∈ P0 (A) implies that |S|f | (F ) − σ| < 1. Let F be a finite subset of a ∈ A : |f (a)| ≥ n1 with at least n(σ + 1) elements. Then X X 1 S|f | (F ∪ F1 ) = |f (a)| ≥ |f (a)| ≥ n(σ + 1) · = σ + 1. n a∈F ∪F1 a∈F But F1 ⊆ F ∪ F1 ∈ P0 (A), so S|f | (F ∪ F1 ) < σ + 1, a contradiction. 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